Problem: Find the gradient of $f(x, y, z) = \tan(xyz)$. $\nabla f = ($ $,$ $,$ $)$
Solution: The gradient of a scalar field is all its partial derivatives put together into a vector. For a 3D scalar field, this looks like $\nabla f = (f_x, f_y, f_z)$. Let's find $f_x$, $f_y$, and $f_z$. We can find $f_x$ by using the chain rule, treating $yz$ as if it were a constant. [What's the derivative of tangent again?] $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ \tan(xyz) \right] \\ \\ &= yz \cdot \dfrac{1}{\cos^2(xyz)} \end{aligned}$ The other two are similar, just replacing the variables that we treat as constant. $\begin{aligned} f_y &= xz \cdot \dfrac{1}{\cos^2(xyz)} \\ \\ f_z &= xy \cdot \dfrac{1}{\cos^2(xyz)} \end{aligned}$ The gradient of $f$ is: $\nabla f = \left( \dfrac{yz}{\cos^2(xyz)}, \dfrac{xz}{\cos^2(xyz)}, \dfrac{xy}{\cos^2(xyz)} \right)$